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Q. Consider the position vector $O P$ of a point $P(x, y, z)$ as in figure shown below
image
If $P_1$ be the foot of perpendicular from the $P$ on the plane $X O Y$, then which of the following options is not true?

Vector Algebra

Solution:

Considering the position vector OP of a point $P(x, y, z)$ as in figure where $P_1$ be the foot of the perpendicular from $P$ on the plane $X O Y$. We, thus, see that $P_1 P$ is
image
parallel to $z$-axis. As $\hat{i}, \hat{j}$ and $\hat{k}$ are the unit vectors along the $x, y$ and $z$-axes, respectively and by the definition of the coordinates of $P$, we have $P_1 P=O R=z \hat{k}$.
Similarly, $Q P_1=O S=y \hat{j}$ and $O Q=x \hat{i}$.
Therefore, it follows that $OP _1= OQ + QP _1=x \hat{ i }+y \hat{ j }$ and $OP = OP _1+ P _1 P =x \hat{i}+y \hat{j}+z \hat{ k }$
Hence, the position vector of $P$ with reference to $O$ is given by
$O P(\text { or } r)=x \hat{i}+y \hat{j}+z \hat{k}$
This form of any vector is called its component form. Here $x$, $y$ and $z$ are called as the scalar components of $r$ and $x \hat{i}, y \hat{j}$ and $z \hat{k}$ are called the vector components of $r$ along the respective axes. Sometimes $x, y$ and $z$ are also termed as rectangular components.
The length of any vector $r=x \hat{i}+y \hat{j}+z \hat{k}$, is readily determined by applying the Pythagoras theorem twice. We note that in the right angle $\triangle O Q P_1$ (fig.)
$\left|O P_1\right|=\sqrt{|O Q|^2+\left|Q P_1\right|^2}=\sqrt{x^2+y^2}$
and in the right angle $\triangle O P_1 P$. we have
$O P=\sqrt{\left|O P_1\right|^2+\left|P_1 P\right|^2}=\sqrt{\left(x^2+y^2\right)+z^2}$
Hence, the length of any vector $r=x \hat{i}+y \hat{j}+z \hat{k}$ is given by
$|r|=|x \hat{i}+y \hat{j}+z \hat{k}|=\sqrt{x^2+y^2+z^2}$
So, only option (d) is incorrect.